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<p><dfn class="terminology">Solution</dfn> We know</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_12.html">
\begin{equation*}
p(x)=1, \quad  q(x)=\frac{1}{1+x^2}.
\end{equation*}
</div>
<p class="continuation">One may observe <span class="process-math">\(p(x)\text{,}\)</span> <span class="process-math">\(q(x)\)</span> are continuous for <span class="process-math">\(-\infty \le x \le \infty\text{.}\)</span> According to the theorem, there is a unique solution which is valid in <span class="process-math">\(-\infty \le x \le \infty\text{.}\)</span> The integrating factor is</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_12.html">
\begin{equation*}
u(x)=e^{\int p(x) \textrm{d} x}=e^{\int  \textrm{d} x}=e^x.
\end{equation*}
</div>
<p class="continuation">Multiplying <span class="process-math">\(u(x)\)</span> on both sides of the ODE, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_12.html">
\begin{equation*}
\label{eq2_12}
\begin{aligned}
&amp; e^x y^{\prime}+e^x y=\frac{e^x}{1+x^2},~\rightarrow~\frac{\textrm{d}}{\textrm{d} x} (e^x y)=\frac{e^x}{1+x^2},\\
&amp;\rightarrow e^x y=\int \frac{e^x}{1+x^2}  \textrm{d} x+C, ~\rightarrow~y=e^{-x}  \left[ \int  \frac{e^x}{1+x^2} \textrm{d} x+C \right].
\end{aligned}
\end{equation*}
</div>
<p class="continuation">Equation (<a href="" class="xref" data-knowl="./knowl/eq2_12.html" title="Equation 2.1.12">(2.1.12)</a>) is the general solution. Then using the initial condition, one has</p>
<div class="displaymath process-math" data-contains-math-knowls="./knowl/eq2_12.html">
\begin{equation*}
\begin{aligned}
&amp;y(0)=0~\rightarrow~0=e^{0}  \left[ \int \frac{e^x}{1+x^2} \textrm{d}  x+C \right]\Big|_{x=0,}\\
&amp;\rightarrow C=-\int \frac{e^x}{1+x^2} \textrm{d}  x\Big|_{x=0.}
\end{aligned}
\end{equation*}
</div>
<span class="incontext"><a href="sec2_2.html#p-26" class="internal">in-context</a></span>
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